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a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
\(1.\)
\(a.\)
\(x^2-2x=x\left(x-2\right)\)
b.
\(3y^3+6xy^2+3x^2y\)
\(=3y\left(y^2+2xy+x^2\right)\)
\(=3y\left(x+y\right)^2\)
\(c.\)
\(x^2-2xy-xy+2y^2\)
\(=x\left(x-2y\right)-y\left(x-2y\right)\)
\(=\left(x-y\right)\left(x-2y\right)\)
\(2.\)
\(a.\)
\(x^2-y^2+5x-5y\)
\(=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+5\right)\)
\(b.\)
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(c.\)
\(x^2-6xy+9y^2-16\)
\(=\left(x^2-6xy+9y^2\right)-4^2\)
\(=\left(x-3\right)^2-4^2\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x-7\right)\left(x+1\right)\)
Tương tự câu \(d,e,g\)
\(3.\)
\(a.\)
\(x^3-2x=0\)
\(\Rightarrow x\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)
\(b.\)
\(x\left(x-4\right)+\left(x-4\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
\(c.\)
\(x\left(x-3\right)+4x-12=0\)
\(\Rightarrow x\left(x-3\right)+3\left(x-3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Tương tự \(d,e,g\)
a) x2( x - 1 ) - x + 1
= x2( x - 1 ) - ( x - 1 )
= ( x - 1 )( x2 - 1 )
= ( x - 1 )( x - 1 )( x + 1 )
= ( x - 1 )2( x + 1 )
b) ( a + b )3 - ( a - b )3
= ( a3 + 3a2b + 3ab2 + b3 ) - ( a3 - 3a2b + 3ab2 - b3 )
= a3 + 3a2b + 3ab2 + b3 - a3 + 3a2b - 3ab2 + b3
= 6a2b + 2b3
= 2b( 3a2 + b )
c) 6x( x - 3 ) + 9 - 3x2
= 6x2 - 18x + 9 - 3x2
= 3x2 - 18x + 9
= 3( x2 - 6x + 3 )
d) x( x - y ) - 5x + 5y
= x( x - y ) - ( 5x - 5y )
= x( x - y ) - 5( x - y )
= ( x - y )( x - 5 )
e) 3( x + 4 ) - x2 - 4x
= 3( x + 4 ) - ( x2 + 4x )
= 3( x + 4 ) - x( x + 4 )
= ( x + 4 )( 3 - x )
f) x2 + 4x - y2 + 4
= ( x2 + 4x + 4 ) - y2
= ( x + 2 )2 - y2
= ( x + 2 - y )( x + 2 + y )
g) x2 + 5x
= x( x + 5 )
h) -x2 + 2x + 2y + y2
= ( y2 - x2 ) + ( 2x + 2y )
= ( y - x )( y + x ) + 2( x + y )
= ( x + y )( y - x + 2 )
3)
e)
b) Ta có: 5x2+10y2-6xy-4x-2y +3= x2 -6xy +(3y)2 +4x2 +y2 -4x -2y +3
= (x - 3y)2 +(2x)2 -4x+1+ y2 -2y+1 +1
= (x-3y)2 + (2x -1)2 + (y-1)2 +1
Ta có :(x-3y)2 luôn lớn hơn hoặc bằng 0
(2x -1)2 luôn lớn hơn hoặc bằng 0
(y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 +1 >0
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
d) D = ( x2 - 4x -5 )(x2 -4x - 19 ) +49
D = ( x2 - 2.2x + 22 -9)(x2 - 2.2x + 22 - 23 ) + 49
D = (x - 2 )2 . ( x-2)2 -9-23+49
D = (x - 2 )2 . ( x-2)2 +17
Do :(x-2)2 lớn hơn hoạc bằng 0 -> (x-2)2 +17 lớn hơn hoặc bằng 17
=> Dmin = 17 khi và chỉ khi : x-2=0 => x=2
Bạn có biết thêm j thì giúp mình nhé :)